coincidental (?) patterns in logs of repeating decimals, e.g. $\ln(2/3)$ vs. $\ln(0.6666666)$

Question

I was playing with logarithms under an arbitrary-precision calculator, and got some odd results. I happened to have the precision set to 20 places, and these are the initial results I got:

ln 0.6666666 = -0.40546520810816938198
ln 2/3       = -0.40546510810816438198

"That's strange", I thought. "I would have expected the logarithms to be more different than that."

But then I looked more closely: they are different, in two places, with the first difference coming at the seventh digit, about as you would expect. But I would have expected all the digits after that to be different — it was quite a surprise that most of them were the same!

To be clear, my assumption is that if two numbers a and b are slightly different, and with a function f such that f(a) and f(b) are also slightly different, then unless there's something special going on, the decimal expansions of f(a) and f(b) will be identical up to a point, then diverge. Unless there's something special going on, I don't expect to see any patterns past the point of divergence — indeed for a transcendental function like ln, I don't expect to see any patterns in a decimal expansion at all. So how can it be that, in this case, there are quite a few digits in common after the point of divergence?

At first I assumed this was just a bizarre coincidence, but then I tried several other numbers, with similar results:

To 13 places:

ln 0.3333333 = -1.0986123886681
ln 1/3       = -1.0986122886681

To 20 places again:

ln 0.1111111 = -2.19722467733622438279
ln 1/9       = -2.19722457733621938279
ln 0.2222222 = -1.50407749677627907337
ln 2/9       = -1.50407739677627407337

(Similar results hold for 4/9, 5/9, 7/9, and 8/9.)

It seems there's got to be "something special" going on here, but I have no idea what! (Although, I realize that patterns in decimal expansions are not necessarily meaningful. I was going to additionally tag this question "numerology", but I guess that's not a tag. :-) )

---

Addendum: The answers reveal that in this case, my assumption that "for a transcendental function like ln, I don't expect to see any patterns in a decimal expansion at all" was quite wrong. The core of the answer to this question is that ln(1 - 10^-7) is −0.000000100000005000000…, and I think this is going to be #3 on my list of fun numerological facts, behind 1001 = 7 · 11 · 13 and, of course, that one about e^iπ.

Also, the answers help to answer two side questions I had but didn't ask at first:

1. Does the effect depend on having truncated 2/3 specifically to 7 digits? No, and it's even more pronounced the more digits you leave in place. For example, if you truncate 2/3 to 15 places, you get an answer that looks like it's correct out to 45 places, typically differing only in the 15th and the 31st.
2. Is there anything special about having 3 or 9 in the denominator? No, the effect usually works for any repeating decimal, as long as (I think) the period is shorter than the truncation point. (Also if the period is greater than 1, you may have to truncate at the right spot, involving multiples of the period.)

Answer 1

For the first example,

$$ 0.6666666 = \frac23 - \frac23 \times 0.0000001 = \frac23 \left(1 - 10^{-7}\right). $$

Therefore $$ \ln(0.6666666) = \ln\left(\frac23\right) + \ln\left(1 - 10^{-7}\right). $$

The Taylor series for the natural logarithm $\ln x$ about $x = 1$ is

$$ \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} -+ \cdots . $$

Therefore \begin{align} \ln\left(1 - 10^{-7}\right) &= -10^{-7} - \frac12\times 10^{-14} - \frac13\times 10^{-21} - \frac14\times 10^{-28} - \cdots \\ &\approx -0.0000001000000050000003333333583333. \end{align}

So you really should expect the difference between the two logarithms to be a difference of $1$ in the seventh digit past the decimal point and a difference of $5$ in the $15$th digit past the decimal point. The next non-zero digits in the difference start at the $22$nd digit past the decimal point.

Depending on what digits are in the smaller logarithm, when you add these differences they may or may not carry into some adjacent digits. If the carries from the $22$nd and later digits don't go very far (and they won't go far unless you have some consecutive $9$s) this means the first $20$ digits are affected only by the $1$ in the seventh digit and the $5$ in the $15$th digit, which is exactly what you see: the seventh digit goes from $1$ to $2$ and the $15$th goes from $4$ to $9.$

The other numbers likewise are in the ratio $1 : \left(1 - 10^{-7}\right),$ so you're adding $1$ to the seventh digit and $5$ to the $15$th digit each time.

Answer 2

Consider the first example, comparing $\log \frac23$ and $\log 0.6666666$. The latter is $\log \left(\frac23 (1 - 10^{-7})\right) = \log \frac23 + \log (1 - 10^{-7})$, so the difference of the logarithms is $-\log (1 - 10^{-7}).$ Expanding $\log (1 - x)$ in a Maclaurin series gives $$\log (1 - x) \sim -x - \frac12 x^2 - \frac13 x^3 + \cdots ,$$ so that difference is a mere $$10^{-7} + \frac12 \cdot 10^{-14} + \cdots = 10^{-7} + 5 \cdot 10^{-15} + \cdots,$$ where $\cdots$ denotes a remainder of $< 10^{-21}$. There turns out to be no carrying, so the first $20$ digits of $\log \frac23$ and $\log 0.6666666$ differ exactly by $1$ in the $7$th place and $5$ in the $15$th place, which is just what you observed.